Question: $g(t)=(t-10)(t+1)$ 1) What are the zeros of the function? Write the smaller $t$ first, and the larger $t$ second. $\text{smaller }t=$
Solution: $\begin{aligned} (t-10)(t+1)&=0 \\\\ t-10=0&\text{ or }t+1=0 \\\\ t={10}&\text{ or }t={-1} \end{aligned}$ There are many ways to find the vertex. We will do it by using the fact that the $t$ -coordinate of the vertex is exactly between the two zeros. $\begin{aligned} \text{vertex's }t\text{-coordinate}&=\dfrac{({10})+({-1})}{2} \\\\ &={\dfrac{9}{2}} \end{aligned}$ Now we can find the vertex's $y$ -coordinate by evaluating $g\left({\dfrac{9}{2}}\right)$ : $\begin{aligned} g\left({\dfrac{9}{2}}\right)&=\left({\dfrac{9}{2}}-10\right)\left({\dfrac{9}{2}}+1\right) \\\\ &=\left(-\dfrac{11}{2}\right)\left(\dfrac{11}{2}\right) \\\\ &=-\dfrac{121}{4} \end{aligned}$ In conclusion, $\begin{aligned} \text{smaller }t&=-1 \\\\ \text{larger }t&=10 \end{aligned}$ The vertex of the parabola is at $\left(\dfrac{9}{2},-\dfrac{121}{4}\right)$